Find $\int \dfrac{1}{4x^2+48x+148}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{4} \text{arctan}\left(\dfrac{x+6}{4}\right)+C$ (Choice B) B $\dfrac{1}{4} \text{arctan}\left(x+6\right)+C$ (Choice C) C $\dfrac{1}{4} \text{arcsin}\left(x+6\right)+C$ (Choice D) D $\dfrac{1}{4} \text{arcsin}\left(\dfrac{x+6}{4}\right)+C$
The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as a constant multiple of $(x+h)^2+k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as a constant multiple of $(x+ h)^2+ k^2$ : $\begin{aligned} 4x^2+48x+148&=4[x^2+12x+37] \\\\ &=4[x^2+12x+36+1] \\\\ &=4[(x+6)^2+1] \\\\ &=4[(x+{6})^2+{1}^2] \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{4x^2+48x+148}\,dx \\\\ &=\int\dfrac{1}{4[(x+6)^2+1^2]}\,dx \\\\ &=\dfrac14\int\dfrac{1}{(x+{6})^2+{1}^2}\,dx \\\\ &=\dfrac14\cdot\dfrac{1}{{1}} \text{arctan}\left(\dfrac{x+{6}}{{1}}\right)+C \\\\ &=\dfrac{1}{4} \text{arctan}\left(x+6\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{4x^2+48x+148}\,dx=\dfrac{1}{4} \text{arctan}\left(x+6\right)+C$